Integrand size = 27, antiderivative size = 89 \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x)) \, dx=\frac {a \operatorname {Hypergeometric2F1}\left (3,\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right ) \sin ^{1+n}(c+d x)}{d (1+n)}+\frac {b \operatorname {Hypergeometric2F1}\left (3,\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin ^{2+n}(c+d x)}{d (2+n)} \] Output:
a*hypergeom([3, 1/2+1/2*n],[3/2+1/2*n],sin(d*x+c)^2)*sin(d*x+c)^(1+n)/d/(1 +n)+b*hypergeom([3, 1+1/2*n],[2+1/2*n],sin(d*x+c)^2)*sin(d*x+c)^(2+n)/d/(2 +n)
Time = 0.12 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00 \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x)) \, dx=\frac {\sin ^{1+n}(c+d x) \left (a (2+n) \operatorname {Hypergeometric2F1}\left (3,\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right )+b (1+n) \operatorname {Hypergeometric2F1}\left (3,\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin (c+d x)\right )}{d (1+n) (2+n)} \] Input:
Integrate[Sec[c + d*x]^5*Sin[c + d*x]^n*(a + b*Sin[c + d*x]),x]
Output:
(Sin[c + d*x]^(1 + n)*(a*(2 + n)*Hypergeometric2F1[3, (1 + n)/2, (3 + n)/2 , Sin[c + d*x]^2] + b*(1 + n)*Hypergeometric2F1[3, (2 + n)/2, (4 + n)/2, S in[c + d*x]^2]*Sin[c + d*x]))/(d*(1 + n)*(2 + n))
Time = 0.31 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3042, 3316, 557, 278}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^n (a+b \sin (c+d x))}{\cos (c+d x)^5}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {b^5 \int \frac {\sin ^n(c+d x) (a+b \sin (c+d x))}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 557 |
\(\displaystyle \frac {b^5 \left (a \int \frac {\sin ^n(c+d x)}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))+b \int \frac {\sin ^{n+1}(c+d x)}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))\right )}{d}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {b^5 \left (\frac {a \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}\left (3,\frac {n+1}{2},\frac {n+3}{2},\sin ^2(c+d x)\right )}{b^5 (n+1)}+\frac {\sin ^{n+2}(c+d x) \operatorname {Hypergeometric2F1}\left (3,\frac {n+2}{2},\frac {n+4}{2},\sin ^2(c+d x)\right )}{b^4 (n+2)}\right )}{d}\) |
Input:
Int[Sec[c + d*x]^5*Sin[c + d*x]^n*(a + b*Sin[c + d*x]),x]
Output:
(b^5*((a*Hypergeometric2F1[3, (1 + n)/2, (3 + n)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(1 + n))/(b^5*(1 + n)) + (Hypergeometric2F1[3, (2 + n)/2, (4 + n)/2 , Sin[c + d*x]^2]*Sin[c + d*x]^(2 + n))/(b^4*(2 + n))))/d
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Sym bol] :> Simp[c Int[(e*x)^m*(a + b*x^2)^p, x], x] + Simp[d/e Int[(e*x)^( m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
\[\int \sec \left (d x +c \right )^{5} \sin \left (d x +c \right )^{n} \left (a +b \sin \left (d x +c \right )\right )d x\]
Input:
int(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c)),x)
Output:
int(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c)),x)
\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x)) \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5} \,d x } \] Input:
integrate(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c)),x, algorithm="fricas" )
Output:
integral((b*sec(d*x + c)^5*sin(d*x + c) + a*sec(d*x + c)^5)*sin(d*x + c)^n , x)
Timed out. \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**5*sin(d*x+c)**n*(a+b*sin(d*x+c)),x)
Output:
Timed out
\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x)) \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5} \,d x } \] Input:
integrate(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c)),x, algorithm="maxima" )
Output:
integrate((b*sin(d*x + c) + a)*sin(d*x + c)^n*sec(d*x + c)^5, x)
Exception generated. \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x)) \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
Exception raised: RuntimeError >> an error occurred running a Giac command :INPUT:sage2OUTPUT:Unable to divide, perhaps due to rounding error%%%{1,[0 ,1,12,0]%%%}+%%%{6,[0,1,10,0]%%%}+%%%{15,[0,1,8,0]%%%}+%%%{20,[0,1,6,0]%%% }+%%%{15,[0,
Timed out. \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x)) \, dx=\int \frac {{\sin \left (c+d\,x\right )}^n\,\left (a+b\,\sin \left (c+d\,x\right )\right )}{{\cos \left (c+d\,x\right )}^5} \,d x \] Input:
int((sin(c + d*x)^n*(a + b*sin(c + d*x)))/cos(c + d*x)^5,x)
Output:
int((sin(c + d*x)^n*(a + b*sin(c + d*x)))/cos(c + d*x)^5, x)
\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x)) \, dx=\left (\int \sin \left (d x +c \right )^{n} \sec \left (d x +c \right )^{5} \sin \left (d x +c \right )d x \right ) b +\left (\int \sin \left (d x +c \right )^{n} \sec \left (d x +c \right )^{5}d x \right ) a \] Input:
int(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c)),x)
Output:
int(sin(c + d*x)**n*sec(c + d*x)**5*sin(c + d*x),x)*b + int(sin(c + d*x)** n*sec(c + d*x)**5,x)*a